\(-2 \times -1\) = \(2\)

taking log to the base ten on both sides we get

\(log -2 + log -1\) = \(log 2\)

\(-log \frac{1}{2} + log -1\) = \(log 2\) SINCE \(log -a\) = \(- log \frac{1}{a}\)

\(log -1\) = \(log 2 + log \frac{1}{2}\)

this can be written as

\(log -1\) = \( log \frac{2}{2}\)

\(log -1\) = \(log 1\)

this implies \(log -1^{3}\) = \(log -1^{2}\) since \(-1^{3}\) = \(-1\) AND \(-1^{2}\) = \(1\)

\(3\) \(log -1\) = \(2\) \(log -1\)

dividing by log -1 to the base ten on both sides,we get

\(3\) = \(2\)

\(3\) = \(1\) + \(1\)

sorry this is a wrong proof

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## Comments

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TopNewestNote that the logarithm formula is \( -\log \frac {1}{a} = \log a\), not "SINCE \( \log -a = - \log \frac {1}{a} \)" as you claimed.

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thanks for correcting my mistake

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i appreicate ur attempt for doing this,i request u think more like this....... (eppadi da room pottu yosipeengaloooo...)-just for fun

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but you're using a number system where 1 plus 1 DOES equal 2 to prove something that contradicts this system... so no.

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